NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion

Ans : (a) Zero net force 
The raindrop is falling with a constant speed. Hence, it acceleration is zero. 
Newton's second law of motion, the net force acting on the rain drop is zero. 
(b) Zero net force 
As per The weight of the cork is acting downward. It is balanced by the buoyant force exerted 
by the water in the upward direction. Hence, no net force is acting on the floating cork. 
(c) Zero net force 
The kite is stationary in the sky, i.e., it is not moving at all. Hence, as per Newton's first 
law of motion, no net force is acting on the kite. 
(d) Zero net force 
The car is moving on a rough road with a constant velocity. Hence, its acceleration is 
zero. As per Newton's second law Of motion, no net force is acting on car. 
(e) Zero net force 
The high speed electron is free from the influence of all fields. Hence, no net force is 
acting on the electron. 

Ans : 0.5 N, in vertically downward direction, in all cases 
Acceleration due to gravity, irrespective of the direction of motion of an object, always 
acts downward. The gravitational force is the only force that acts on the pebble in all 
three cases. Its magnitude is given by Newton's second law of motion as: 
\(\begin{array}{l}{F=m \times a} \\ {\text { Where, }} \\ {F=\text { Net force }} \\ {m=\text { Mass of the pebble }=0.05 \mathrm{kg}} \\ {a=\mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2}} \\ {\therefore F=0.05 \times 10=0.5 \mathrm{N}}\end{array}\)
The net force on the pebble in all three cases is 0.5 N and this force acts in the 
downward direction. 
If the pebble is thrown at an of 450 with the horizontal, it will have both the 
horizontal and vertical components Of velocity. At the highest point, only the vertical 
component of velocity becomes zero. However, the pebble will have the horizontal 
component of velocity throughout its motion. This component of velocity produces no 
effect on the net force acting on the pebble. 

Ans : (a) l N; vertically downward 
Mass of the stone, m 0.1 kg 
Acceleration Of the stone, a = g = 10 \(\mathrm{m} / \mathrm{s}^{2}\) 
As per Newton's second law of motion, the net force acting on the stone, 
F ma = mg 
=0.1 x 10=1 N 
(b) l N; vertically downward 
The train is moving with-I a constant velocity. Hence, its acceleration is zero in the 
direction of its motion, i.e., in tie horizontal direction. Hence, no force is acting on ene 
stone in the horizontal direction. 
The net force acting on the stone is because of acceleration due to gravity and it always 
acts vertically downward. The magnitude of this force is 1 N.
(c) l N; vertically downward 
It is given that the train is accelerating at the rate Of 1 \(\mathrm{m} / \mathrm{s}^{2}\) 
Therefore, the net force acting on the stone, F' = ma — -0.1 x 1-0.1 N 
This force is acting in the horizontal direction. Now, when the stone is dropped, the 
horizontal force F,' stops acting on the stone. This is because of the fact that the force 
acting on a body at an instant depends on the situation at that instant and not on earlier 
Situations.
Therefore, the net force acting on the stone is given only by acceleration due to gravity. 
\(F=m g=1 \mathrm{N}\)
This force acts vertically downward.  
(d) O.1 N, in the direction of motion of the train 
The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train. 
Acceleration of the train, a = 0.1 \(\mathrm{m} / \mathrm{s}^{2}\)  
The net force acting on the stone will be in the direction of motion of the train. Its 
magnitude is given by: 
F = ma 
-0.1 x 1-0.1 N 

Ans : (i)When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T, i.e 
\(F=T=\frac{m v^{2}}{l}\)
Where F is the net force acting on the particle. 

Ans : Retarding force, F = —50 N 
Mass of the body, m = 20 kg 
Initial velocity of the body, u = 15 m/s 
Final velocity Of the body, V = O 
using Newton's second law of motion, the acceleration (a) produced in the body can be 
calculated as: 
\(\begin{array}{l}{F=m a} \\ {-50=20 \times a} \\ {\therefore a=\frac{-50}{20}=-2.5 \mathrm{m} / \mathrm{s}^{2}}\end{array}\)
using the first equation Of motion, the time (t) taken by the body to come to rest can be 
calculated as: 
\(\begin{array}{l}{v=u+a t} \\ {\therefore t=\frac{-u}{a}=\frac{-15}{-2.5}=6 \mathrm{s}}\end{array}\)